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Suppose $|G|=p^nm$ with $1<m<p$. Show that $G$ is not simple.

I know it has to do with group actions. My idea is to consider a subgroup of order $p^n$, call it $H$. It has index $m$, so there are $m$ left cosets of $H$ in $G$, and $S_m$ acts on the set of left cosets. This gives a group homo $G\to S_m$. How do I proceed? I guess either $H$ should be normal or I should prove that the kernel of this map is a proper nontrivial subgroup, which is normal. But I don't know how to do either of those.

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Sylow's theorem is the easiest way to go here. The number of $p$-Sylow subgroups is congruent to $1\pmod p$ and also divides $m$. Since $m<p$ there must be only one $p$-Sylow subgroup, which is therefore normal. Thus $G$ is not simple because it has a normal subgroup of order $p^n$.

Or, to finish your proof, it is true that $G$ acts on the cosets of $H$. So there is an homomorphism from $G$ to $S_m$ which cannot be an isomorphism since $p$ does not divide $m!$, which is the order of $S_m$. Also, it can't map all of $G$ to $1\in S_m$ since the action on cosets is transitive. Thus, the kernel is a non-trivial normal subgroup, and $G$ is not simple.

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You can use your idea to finish the proof. The kernel of the map is nontrivial, since $p$ does not divide the order of $S_m$.

Or you could use Sylow's theorems to prove that $H$ is normal.

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